Quant puzzles, solved step by step

Puzzles
100
Traced to a firm
71
Topics
12
Hard
30
Topic
All topicsLogic and algorithmic reasoning10Conditional probability and Bayes7Counting and combinatorics8Continuous and geometric probability9Correlation, regression and linear algebra9Market making, betting and sizing9Expected value and optimal stopping9Statistics and estimation9Pricing, options and index maths7Games and strategic reasoning8Markov chains and random walks7Mental maths and number sense8
Level
AnyWarm upCoreHard
Source
AnyReported at a firmStandard
Showing 1–2 of 2 · filtered from 100Clear filters
  1. 008Z1 and Z2 are independent standard normal random variables. What are the mean and variance of Z1 squared + Z2 squared, and what is the probability that it exceeds 2?Statistics and estimationCoreQuant researchQuant trading

    Try it first

    What is P(Z1 squared + Z2 squared > 2)?

    Show the worked solution

    Mean 2, variance 4, and the probability of exceeding 2 is e to the -1, about 36.8%. Each Z squared has mean 1 and variance 2, so the sum has mean 2 and variance 4. The sum is a chi-squared with two degrees of freedom, which happens to be exactly an exponential with mean 2. Its tail beyond t is e to the -t/2, so beyond 2 it is e to the -1.

    How do you get the mean and variance without the distribution?

    E[Z squared] is the variance of Z, which is 1. For the variance of Z squared you need the fourth moment: E[Z to the 4] is 3 for a standard normal. So Var(Z squared) = 3 - 1 = 2, and because Z1 and Z2 are independent the variances add, giving a mean of 2 and a variance of 4. Say the fourth moment of 3 aloud; it is the number interviewers check you know, and it is why normal kurtosis is quoted as 3.

    Two squared normals add up to an exponential with mean 2024680.250.5mean = 263.2%P(above 2) = e to the -1 = 36.8%value of Z1 squared + Z2 squaredDensity: (1/2) e to the -x/2Mean 2, variance 4Tail: P(X > t) = e to the -t/2
    The sum of two squared standard normals has density one half times e to the minus x over 2, an exponential with mean 2 and variance 4, and the area beyond 2 is exactly e to the minus 1, about 36.8%.

    Why is this particular sum exponential?

    Think of a dart thrown at a board where both the horizontal and vertical errors are independent standard normals. The joint density depends only on the distance from the centre, so the dart's direction is uniform and all the information is in the radiusThe distance of the point (Z1, Z2) from the origin, the square root of Z1 squared plus Z2 squared.. Switching to polar coordinates, the chance that the squared distance exceeds t is e to the -t/2, which is the tail of an exponential with mean 2. At t = 2 the answer is e to the -1.

    The relationship
    P(Z12+Z22>t)=∫02π ⁣ ⁣∫t∞12πe−r2/2 r dr dθ=e−t/2P( ⋅>2)=e−1P(Z_1^2+Z_2^2 > t) = \int_0^{2\pi}\!\!\int_{\sqrt t}^{\infty} \frac{1}{2\pi} e^{-r^2/2}\, r\,dr\,d\theta = e^{-t/2} \qquad P(\,\cdot > 2) = e^{-1}
    rthe distance of (Z1, Z2) from the origin
    \frac{1}{2\pi} e^{-r^2/2}the joint density of two independent standard normals
    e^{-t/2}the tail of an exponential distribution with mean 2
    What it says in wordsIn polar coordinates the angle integrates out and the radius gives an exponential tail.

    Why is 50% the tempting wrong answer?

    Because 2 is the mean and people read the mean as the middle. For a right-skewed distribution the mean sits above the median, so less than half the mass lies beyond it; here the median is 2 ln 2, about 1.39. A simulation of 200,000 pairs gives a mean of 1.997, a variance of 3.96 and a tail share of 0.368, matching the exact results. The same polar trick is what powers the Box-Muller method for generating normal random numbers.

    Where candidates lose it

    The common wrong answer is 50%, from treating the mean as the median. Chi-squared variables are skewed to the right, and the skew is largest with few degrees of freedom.

    The second trap is the variance. Candidates who say the variance of Z squared is 1 have confused it with the variance of Z. The fourth moment of 3 is the step, and missing it gives a variance of 2 for the sum instead of 4.

    What the interviewer asks next

    • What is the distribution of the square root of Z1 squared + Z2 squared?
    • How would you use this to generate normal random numbers from uniforms?
    • What are the mean and variance of a chi-squared with k degrees of freedom?
  2. 057Z is a standard normal random variable. What is the expected value of max(Z, 0), and what does that number tell you about the price of an at-the-money option?Statistics and estimationCoreQuant researchQuant trading

    Try it first

    Roughly what is E[max(Z, 0)]?

    Show the worked solution

    1/sqrt(2 pi), about 0.399. Only the positive half contributes, and there you integrate z times the normal density. Because the derivative of the density is minus z times the density, the integral is simply the density's height at zero. So an at-the-money call on a normally distributed move is worth about 0.4 standard deviations of that move: roughly 0.4 x S x volatility x the square root of time.

    Why is the answer not zero, and not one half?

    Think of a shop that keeps the profit on good days and closes, losing nothing, on bad days. Its average day is better than the average of all days, because the bad days have been floored. max(Z, 0) throws away every negative outcome and keeps every positive one at its full size, so its mean is the positive half's contribution alone: the integral of z times the density from zero to infinity. That is not one half, which is only the chance of being positive; the size of each positive draw matters too.

    Only the right half pays, and it pays z: the area is 0.399-3-2-10123z, in standard deviationspeak height 1/sqrt(2 pi) = 0.399density of Zz x density, z > 0area = 0.399negative side pays 0E[max(Z, 0)]= 0.399At the money call:0.4 x S x vol x sqrt(T)S 1,000, vol 20%, 3 months0.399 x 1,000 x 0.2 x 0.5 = 39.9Black-Scholes, zero rates: 39.88
    The shaded area under z times the normal density on the positive side is exactly 0.399, the same as the density's peak height, so an at-the-money call on a normal move of one standard deviation is worth about 0.4, which for a Rs 1,000 stock at 20% volatility over three months is about Rs 40.

    How do you do the integral in one line?

    Notice what differentiating the density gives. The derivative of e to the minus z squared over 2 is minus z times itself, so z times the density is the negative derivative of the density, and its integral from 0 to infinity is the density at 0 minus the density at infinity. The density at infinity is zero, and at zero it is 1/sqrt(2 pi). No tables, no substitution: the answer is 0.3989. Doubling it gives E[|Z|], about 0.798, which is the at-the-money straddle.

    The relationship
    E[max⁡(Z,0)]=∫0∞z φ(z) dz=[−φ(z)]0∞=φ(0)=12π≈0.399E[\max(Z,0)] = \int_0^\infty z\,\varphi(z)\,dz = \big[-\varphi(z)\big]_0^\infty = \varphi(0) = \frac{1}{\sqrt{2\pi}} \approx 0.399
    phi(z)the standard normal density, e^(-z^2/2) / sqrt(2 pi)
    phi(0)the height of the density at its peak
    What it says in wordsThe expected positive part of a standard normal equals the height of the bell at its centre.

    What does it say about an at-the-money option?

    If the stock's move to expiry is roughly normal with standard deviation S x vol x sqrt(T), an at-the-money call pays the positive part of that move. So its value is about 0.4 x S x vol x sqrt(T), the rule of thumb option traders use to price at-the-money options in their heads. For a Rs 1,000 stock at 20% volatility and three months, sqrt(T) is 0.5 and the call is about 0.399 x 1,000 x 0.2 x 0.5 = Rs 39.9; the Black-Scholes value with zero rates is Rs 39.88. The rule loosens for long maturities and high volatilities, where the lognormal skew matters.

    Where candidates lose it

    The two fast wrong answers are 0, from averaging Z itself, and 0.5, from confusing the probability of a positive draw with its expected size. Both come from answering before writing down what is being averaged.

    The second loss is getting 0.399 and not connecting it to options, which is why the question is asked on a trading desk. Say the 0.4 rule in the same breath.

    What the interviewer asks next

    • What is E[max(Z, 1)]?
    • What is the variance of max(Z, 0)?
    • Using the rule, what is an at-the-money straddle worth on a Rs 500 stock at 30% volatility for one month?